Re: [中學] 無窮級數
※ 引述《susiseptem (..)》之銘言：
: 無窮級數 1 至 無窮大 n(1/3)^n 之值
: 感謝板上專業的大大QQ
: 100P答謝
∞ n 2 3
令 F = Σ n(1/3) = 1/3 +2(1/3) + 3(1/3) + ....
n=1
同乘1/3
2 3
=>F/3 = (1/3) + 2(1/3) + .....
2 3 ∞ n 1/3
=> F-F/3 = 1/3 + (1/3) + (1/3) + ... = Σ (1/3) = ------- = 1/2
n=1 1-1/3
=> F = 3/4
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05/15 23:58, , 1F
05/15 23:58, 1F
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